How Do Protons and Alpha Particles Behave After an Elastic Collision?

[bigtymer8700]

Collisions between atomic and subatomic particles are usually perfectly elastic. In one such collision, a proton traveling to the right at 274 km/s collides elastically with a stationary alpha particle (a helium nucleus, having mass 6.65×10−27 kg . I am trying to figure out the velocity of after collision for helium nucleus

Ma=?
Mb=6.65 x10-27kg
Vai= 274000m/s


i started off using the cons. of momentum so i got

Ma(274000m/s)+6.65^-27 kg(0 m/s) = MaVaf + Mb+Vbf

i know I am missing a step here to continue working the problem i just can't figure out where

 

[Astronuc]

MaVaf + Mb+Vbf

should be MaVaf + MbVbf, but I presume that is a typo.

Well - one has two unknowns Vaf and Vbf, so one needs two equations.

One is conservation of momentum and the other is conservation of energy.

Now, is one assuming a head on collision without any lateral deflection?

See - http://hyperphysics.phy-astr.gsu.edu/hbase/colsta.html

But look at Rutherford scattering -

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html#c4
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/ruthlink.html

 

[bigtymer8700]

yes its a head on collision no lateral deflection. so how would you use the conservation of energy to solve for the unknown of the two equations?

 

[Astronuc]

Kinetic energy before = kinetic energy after, similar to conservation of momentum.

274 km/s = 0.000913c, so non-relativistic mechanics is reasonable.

 

[bigtymer8700]

yes so would you then set up 1/2MaVai^2 +1/2MbVbi^2=1/2MaVaf^2+ 1/2MbVbf^2 to get the unknown Mass or velocity?

 

[Astronuc]

Yes. One knows the mass of a proton and helium atom (nucleus = alpha particle). Assume the helium atom is stationary, which it should be, otherwise calculate its velocity assuming room temperature.

 

[bigtymer8700]

thats where I am stuck on i have both equations set up but i don't see what i have that can be substituted in for each other. i know its somewhere simple

I have 1/2Ma(274000m/s)^2+ 0 = 1/2MaVaf^2 + 1/2 6.65^-27kg Vbf^2

thats for the conservation of KE

you already see what i have for the cons. of momentum up there i just can't see what unknown i have to isolate first to then solve for it

 

[Astronuc]

Start with what one knows.

Besides the mass of the He nucleus (6.65 E-27 kg), one knows the mass of a proton, which is 1.67 E-27 kg. The proton is moving at 2.74 E5 m/s.

One can compute the momentum (mv) and kinetic energy (1/2mv²) of the proton. The values are 0 for the He nucleus.

Momentum eq: m_pv_pi= m_pv_pf + m_av_af, where m_p is the mass of the proton and m_a is the mass of the alpha particle, and i,f subscripts denote initial and final quantities.

Now one could divide the momentum equation by m_p and obtain an equation of one unknown velocity in terms of the other, then substitute that into the energy equation.

Energy eq: m_pv_pi²= m_pv_pf² + m_av_af² (the 1/2 factored out)

Similarly, one could divide by m_p.

The quantities on the left-hand sides of the momentum and energy equations are known.